To find the unit-vector notation for vector $\mathbf{G}$, given that it has a magnitude of 31.0 and its direction is 40.0° clockwise from the negative $y$-axis, we follow these steps:

Step 1: Understand the direction

• The vector is 40.0° clockwise from the negative $y$-axis.
• If we convert this to the standard coordinate system (where angles are measured counterclockwise from the positive $x$-axis), we subtract 40.0° from 270° (the angle for the negative $y$-axis): $\theta = 270^\circ - 40^\circ = 230^\circ$ So the angle of the vector from the positive $x$-axis is 230°.

Step 2: Break the vector into components

• The magnitude of the vector is $|\mathbf{G}| = 31.0$.
• The vector components in terms of the angle $\theta$ are: $G_x = |\mathbf{G}| \cdot \cos(\theta)$ $G_y = |\mathbf{G}| \cdot \sin(\theta)$

Substitute $|\mathbf{G}| = 31.0$ and $\theta = 230^\circ$: $G_x = 31.0 \cdot \cos(230^\circ)$ $G_y = 31.0 \cdot \sin(230^\circ)$

Step 3: Calculate the components

Using a calculator: $G_x = 31.0 \cdot \cos(230^\circ) \approx 31.0 \cdot (-0.6428) = -19.9$ $G_y = 31.0 \cdot \sin(230^\circ) \approx 31.0 \cdot (-0.7660) = -23.7$

Step 4: Express in unit-vector notation

Thus, the vector $\mathbf{G}$ in unit-vector notation is: $\mathbf{G} = -19.9 \, \hat{i} - 23.7 \, \hat{j}$ where $\hat{i}$ is the unit vector in the $x$-direction, and $\hat{j}$ is the unit vector in the $y$-direction.

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5 Related Questions:

1. How do you find the unit-vector notation of a vector given its magnitude and direction in general?
2. What is the significance of clockwise vs. counterclockwise angles in vector problems?
3. How can you convert between angle measurements from different reference axes?
4. How does vector addition work in unit-vector notation?
5. What are the properties of unit vectors, and how do they simplify vector calculations?

Tip:

When dealing with angles, always double-check the reference direction (clockwise or counterclockwise) to avoid sign mistakes in your calculations.